focal length of the curve C2 is

Engineering

Mathematics

Standard Ellipse

Question

focal length of the curve C₂ is

$2 \sqrt{3}$

$4 \sqrt{6}$

$4 \sqrt{3}$

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y² = 4ax (k = 4a)

$P (λ, λ) \equiv P (4 a, 4 a) \{∵ {at}^{2} = 2 at \Rightarrow t = 2\}$

Tangent at P
y · 4a = 2a (x + 4a)
2y = x + 4a
T ≡ (0, 2a)

$PT = 4 \sqrt{5} \Rightarrow 16 a^{2} + 4 a^{2} = 80 \Rightarrow a = 2$

∴ y² = 8x

$C_{2} : \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$

$e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

$focal length, 2 ae = 2 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 4 \sqrt{3}$